How do you solve $6-p^2/8=-4$ ?

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Answer 1 · 2016-10-30T09:41:26

$p=+-4sqrt5$

$6-p^2/8=-4$

$hArr6+4=p^2/8$

or $p^2/8=10$

or $p^2=80$

or $p^2-80=0$

or $p^2-(sqrt80)^2=0$

or $(p+sqrt80)(p-sqrt80)=0$

i.e. either $p+sqrt80=0$ or $p-sqrt80=0$

i.e. either $p=-sqrt80$ or $p=sqrt80$

i.e. $p=+-sqrt(4×4×5)=+-4sqrt5$

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